Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}9x+2y &= 8 \\ 3x-2y &= 4\end{align*}$
Explanation: Begin by moving the $y$ -term in the second equation to the right side of the equation. $3x = 2y+4$ Divide both sides by $3$ to isolate $x$ $x = {\dfrac{2}{3}y + \dfrac{4}{3}}$ Substitute this expression for $x$ in the first equation. $9({\dfrac{2}{3}y + \dfrac{4}{3}}) + 2y = 8$ $6y + 12 + 2y = 8$ Simplify by combining terms, then solve for $y$ $8y + 12 = 8$ $8y = -4$ $y = -\dfrac{1}{2}$ Substitute $-\dfrac{1}{2}$ for $y$ in the top equation. $9x+2( -\dfrac{1}{2}) = 8$ $9x-1 = 8$ $9x = 9$ $x = 1$ The solution is $\enspace x = 1, \enspace y = -\dfrac{1}{2}$.